I will solve just one specific case:
For this case, we need to solve $x^2 \equiv -125 \pmod{22}$
We have $22|(x^2+125)\Rightarrow22|(x^2-7+132)\Rightarrow22|(x^2-7)$ because $132$ is divisible by $22$.
So now we need to solve $x^2 \equiv 7 \pmod{22}$
Trial and error is actually suitable, but only if you notice that because $x^2-7$ is divisible by $22$, $x$ must be an odd number. This will reduce the number of cases from $22$ to $11$. With $11$ cases, we can easily set up a table like this:
\begin{array}{|c|c|c|c|c|c|c|c|c|} \hline \text{Remainder of }x\div22 &1&3&5&7&9&11&13&15&17&19&21\ \\ \hline \text{Remainder of }x^2\div22 &1&9&3&5&15&11&15&5&3&9&1\\\hline \text{Remainder of }(x^2-7)\div22 &16&2&18&20&8&4&8&20&18&2&16\\\hline \end{array}
Based on the table above, there are no positive integers $x$ satisfy $x^2 \equiv 7 \pmod{22}$, so there are no positive integers $x,y$ overall satisfy $125+x^2=22y$.